Maths-

General

Easy

Question

# A 12% brine solution was mixed with a 16% brine solution to produce a 15% brine solution. How much of the 12% brine solution and how much of the 16% brine solution were used to produce 40 L of the 15% solution?

Hint:

### let the 12 % brine solution be x L and the 16% brine solution mixed be y L

Total amount of mixed solution = amount of 12 % solution +amount of 16 % solution

= 40 L

Amount of salt in final brine solution = amount of salt in 15% brine solution

= amount of salt in 12% brine + amount of salt in 16% brine

## The correct answer is: 16%

### Ans :- 30 L of 12% brine solution and 10 L of 16% brine solution

Explanation :-

Step 1:-Frame system of linear equations by given condition

Total amount of mixed solution = amount of 12 % solution +amount of 16 % solution

— Eq1

The amount of salt in 12% of x Litres brine solution =

The amount of salt in 16% of y Litres brine solution =

The amount of salt in 15% of 40 L of brine solution =

Amount of salt in final brine solution = amount of salt in 15% brine solution

= amount of salt in 12% brine + amount of salt in 16% brine

We get, — Eq2

Step 2:- eliminate x to find the value of y .

Doing (Eq2) -3(Eq1) to eliminate x .

∴y = 30

Step 3:- find x by substituting the value of y in eq1.

∴x =10

∴30 L of 12% brine solution and 10 L of 16% brine solution is mixed to get 40 L of 15 % brine solution .

### Related Questions to study

Maths-

### Candies cost $2 an ounce and have 500 calories in an ounce. Cookies cost $3 an ounce and have 400calories in an ounce. A mixture of candies and cookies costs $29 and has 4,800 calories. Create a system of equations to model this scenario. How many ounces of candies are there in the mixture?

Ans :-The no. of ounces of candy is 4 ounces in the mixture .

Explanation :-

Given ,candies cost $2 an ounce and have 500 calories in an ounce. Cookies cost $3 an ounce and have 400calories in an ounce.

Step 1:Frame the system of linear equations based on given conditions

Let the no. of ounces of candy be x and no. of ounces of cookies be y

Total cost = cost of x ounces candy + cost of y ounces cookies =

x cost of each ounce candy + y cost of each ounce cookies = $29

— Eq1

Total no. of calories = calories in x ounces candy + calories in y ounces cookies =

x calories in each ounce candy + y calories in each ounce cookies = 4,800

— Eq2

Step 2:- eliminate x to find the value of y .

Doing 5(Eq1) -2(Eq2) to eliminate x .

Step 3:- find x by substituting the value of y in eq1.

∴ we get the no. of ounces of candy be x = 4 ounces.

Explanation :-

Given ,candies cost $2 an ounce and have 500 calories in an ounce. Cookies cost $3 an ounce and have 400calories in an ounce.

Step 1:Frame the system of linear equations based on given conditions

Let the no. of ounces of candy be x and no. of ounces of cookies be y

Total cost = cost of x ounces candy + cost of y ounces cookies =

x cost of each ounce candy + y cost of each ounce cookies = $29

— Eq1

Total no. of calories = calories in x ounces candy + calories in y ounces cookies =

x calories in each ounce candy + y calories in each ounce cookies = 4,800

— Eq2

Step 2:- eliminate x to find the value of y .

Doing 5(Eq1) -2(Eq2) to eliminate x .

∴y = 7 |

∴x =4 |

### Candies cost $2 an ounce and have 500 calories in an ounce. Cookies cost $3 an ounce and have 400calories in an ounce. A mixture of candies and cookies costs $29 and has 4,800 calories. Create a system of equations to model this scenario. How many ounces of candies are there in the mixture?

Maths-General

Ans :-The no. of ounces of candy is 4 ounces in the mixture .

Explanation :-

Given ,candies cost $2 an ounce and have 500 calories in an ounce. Cookies cost $3 an ounce and have 400calories in an ounce.

Step 1:Frame the system of linear equations based on given conditions

Let the no. of ounces of candy be x and no. of ounces of cookies be y

Total cost = cost of x ounces candy + cost of y ounces cookies =

x cost of each ounce candy + y cost of each ounce cookies = $29

— Eq1

Total no. of calories = calories in x ounces candy + calories in y ounces cookies =

x calories in each ounce candy + y calories in each ounce cookies = 4,800

— Eq2

Step 2:- eliminate x to find the value of y .

Doing 5(Eq1) -2(Eq2) to eliminate x .

Step 3:- find x by substituting the value of y in eq1.

∴ we get the no. of ounces of candy be x = 4 ounces.

Explanation :-

Given ,candies cost $2 an ounce and have 500 calories in an ounce. Cookies cost $3 an ounce and have 400calories in an ounce.

Step 1:Frame the system of linear equations based on given conditions

Let the no. of ounces of candy be x and no. of ounces of cookies be y

Total cost = cost of x ounces candy + cost of y ounces cookies =

x cost of each ounce candy + y cost of each ounce cookies = $29

— Eq1

Total no. of calories = calories in x ounces candy + calories in y ounces cookies =

x calories in each ounce candy + y calories in each ounce cookies = 4,800

— Eq2

Step 2:- eliminate x to find the value of y .

Doing 5(Eq1) -2(Eq2) to eliminate x .

∴y = 7 |

∴x =4 |

Maths-

### Chocolate costs $7 a pound. Biscuits cost $9 a pound. There is a mixture of peanuts and cashews that weighs 10pounds and costs $84. How many pounds of chocolates are there in the mixture?

Ans :- weight of chocolate is 3 (Pounds) in the mixture of chocolates and biscuits of 10 pounds.

Explanation :-

Chocolate cost $7 a pound. Biscuits cost $9 a pound

Step 1:- frame the system of linear equations with given conditions.

let the weight of chocolate be x (Pounds) and the weight of biscuits be y(Pounds).

Given total weight = weight of chocolates + weight of biscuits = 10(Pounds)

x + y = 10 — eq1

Given total cost of 10 pound mixture = cost of x pounds chocolate + cost of y pounds Biscuits = $94.

Cost of x pounds chocolate = x cost of chocolate for 1 pound( = $7 given)

Cost of x pounds chocolate = x $7 = 7x(in $)

Cost of y pounds of biscuits = y cost of biscuits for 1 pound (= $9 given )

Cost of y pounds of biscuits = y $9 = 9y (in $)

So total cost of 10 pounds mixture = — eq2

Step 2:- eliminate the x and find y

Doing eq2 - 3(eq1) to eliminate x .

Step 3:- find x by substituting the value of y in eq1.

∴weight of chocolate is 3 (Pounds) in the mixture of chocolates and biscuits of 10 pounds.

Explanation :-

Chocolate cost $7 a pound. Biscuits cost $9 a pound

Step 1:- frame the system of linear equations with given conditions.

let the weight of chocolate be x (Pounds) and the weight of biscuits be y(Pounds).

Given total weight = weight of chocolates + weight of biscuits = 10(Pounds)

x + y = 10 — eq1

Given total cost of 10 pound mixture = cost of x pounds chocolate + cost of y pounds Biscuits = $94.

Cost of x pounds chocolate = x cost of chocolate for 1 pound( = $7 given)

Cost of x pounds chocolate = x $7 = 7x(in $)

Cost of y pounds of biscuits = y cost of biscuits for 1 pound (= $9 given )

Cost of y pounds of biscuits = y $9 = 9y (in $)

So total cost of 10 pounds mixture = — eq2

Step 2:- eliminate the x and find y

Doing eq2 - 3(eq1) to eliminate x .

∴y = 7 |

∴x = 3 |

### Chocolate costs $7 a pound. Biscuits cost $9 a pound. There is a mixture of peanuts and cashews that weighs 10pounds and costs $84. How many pounds of chocolates are there in the mixture?

Maths-General

Ans :- weight of chocolate is 3 (Pounds) in the mixture of chocolates and biscuits of 10 pounds.

Explanation :-

Chocolate cost $7 a pound. Biscuits cost $9 a pound

Step 1:- frame the system of linear equations with given conditions.

let the weight of chocolate be x (Pounds) and the weight of biscuits be y(Pounds).

Given total weight = weight of chocolates + weight of biscuits = 10(Pounds)

x + y = 10 — eq1

Given total cost of 10 pound mixture = cost of x pounds chocolate + cost of y pounds Biscuits = $94.

Cost of x pounds chocolate = x cost of chocolate for 1 pound( = $7 given)

Cost of x pounds chocolate = x $7 = 7x(in $)

Cost of y pounds of biscuits = y cost of biscuits for 1 pound (= $9 given )

Cost of y pounds of biscuits = y $9 = 9y (in $)

So total cost of 10 pounds mixture = — eq2

Step 2:- eliminate the x and find y

Doing eq2 - 3(eq1) to eliminate x .

Step 3:- find x by substituting the value of y in eq1.

∴weight of chocolate is 3 (Pounds) in the mixture of chocolates and biscuits of 10 pounds.

Explanation :-

Chocolate cost $7 a pound. Biscuits cost $9 a pound

Step 1:- frame the system of linear equations with given conditions.

let the weight of chocolate be x (Pounds) and the weight of biscuits be y(Pounds).

Given total weight = weight of chocolates + weight of biscuits = 10(Pounds)

x + y = 10 — eq1

Given total cost of 10 pound mixture = cost of x pounds chocolate + cost of y pounds Biscuits = $94.

Cost of x pounds chocolate = x cost of chocolate for 1 pound( = $7 given)

Cost of x pounds chocolate = x $7 = 7x(in $)

Cost of y pounds of biscuits = y cost of biscuits for 1 pound (= $9 given )

Cost of y pounds of biscuits = y $9 = 9y (in $)

So total cost of 10 pounds mixture = — eq2

Step 2:- eliminate the x and find y

Doing eq2 - 3(eq1) to eliminate x .

∴y = 7 |

∴x = 3 |

Maths-

### A fence is put around a rectangular plot of land. The perimeter of the fence is 28 feet. Two of the opposite sides of the fence cost $10 per foot. The other two sides cost $12 per foot. If the total cost of the fence is $148, what are the dimensions of the fence?

Ans :- 10 by 4 is the dimensions of the rectangular plot fence

Explanation :-

Step 1:- frame the system of linear equations from given conditions

let the length of plot be 1 and let the breadth of plot be b

Given total perimeter of the plot = 2(1 + b) = 28 — eq1

let the breadth b fence cost $12 per foot .

let the length 1 fence cost $10 per foot

Given total cost of fence = cost of fencing length 1 + cost of fencing breadth b

cost of fencing length 1 = 1 cost of fencing per foot .

cost of fencing length 1 = $10(1)

cost of fencing breadth b = b cost of fencing per foot

cost of fencing breadth b = $12(b)

Total cost of fencing be = 10(1) + 12b

10l + 12b = 148 5l + 6b = 74 — eq 2

Step 2 :- eliminate x to find value of y

Doing 5eq1 - eq2 to eliminate x

We get 5l + 5b = (5l + 6b) = 5(14) = 74

5l + 5b - 5l - 6b = 70 - 74

-b = -4

Step 3:- substitute the value of y in eq1 to get the value of x

∴ we get the dimensions of rectangular plot = 10 by 4 (in foots)

Explanation :-

Step 1:- frame the system of linear equations from given conditions

let the length of plot be 1 and let the breadth of plot be b

Given total perimeter of the plot = 2(1 + b) = 28 — eq1

let the breadth b fence cost $12 per foot .

let the length 1 fence cost $10 per foot

Given total cost of fence = cost of fencing length 1 + cost of fencing breadth b

cost of fencing length 1 = 1 cost of fencing per foot .

cost of fencing length 1 = $10(1)

cost of fencing breadth b = b cost of fencing per foot

cost of fencing breadth b = $12(b)

Total cost of fencing be = 10(1) + 12b

10l + 12b = 148 5l + 6b = 74 — eq 2

Step 2 :- eliminate x to find value of y

Doing 5eq1 - eq2 to eliminate x

We get 5l + 5b = (5l + 6b) = 5(14) = 74

5l + 5b - 5l - 6b = 70 - 74

-b = -4

∴ b = 4 |

∴ l = 10. |

### A fence is put around a rectangular plot of land. The perimeter of the fence is 28 feet. Two of the opposite sides of the fence cost $10 per foot. The other two sides cost $12 per foot. If the total cost of the fence is $148, what are the dimensions of the fence?

Maths-General

Ans :- 10 by 4 is the dimensions of the rectangular plot fence

Explanation :-

Step 1:- frame the system of linear equations from given conditions

let the length of plot be 1 and let the breadth of plot be b

Given total perimeter of the plot = 2(1 + b) = 28 — eq1

let the breadth b fence cost $12 per foot .

let the length 1 fence cost $10 per foot

Given total cost of fence = cost of fencing length 1 + cost of fencing breadth b

cost of fencing length 1 = 1 cost of fencing per foot .

cost of fencing length 1 = $10(1)

cost of fencing breadth b = b cost of fencing per foot

cost of fencing breadth b = $12(b)

Total cost of fencing be = 10(1) + 12b

10l + 12b = 148 5l + 6b = 74 — eq 2

Step 2 :- eliminate x to find value of y

Doing 5eq1 - eq2 to eliminate x

We get 5l + 5b = (5l + 6b) = 5(14) = 74

5l + 5b - 5l - 6b = 70 - 74

-b = -4

Step 3:- substitute the value of y in eq1 to get the value of x

∴ we get the dimensions of rectangular plot = 10 by 4 (in foots)

Explanation :-

Step 1:- frame the system of linear equations from given conditions

let the length of plot be 1 and let the breadth of plot be b

Given total perimeter of the plot = 2(1 + b) = 28 — eq1

let the breadth b fence cost $12 per foot .

let the length 1 fence cost $10 per foot

Given total cost of fence = cost of fencing length 1 + cost of fencing breadth b

cost of fencing length 1 = 1 cost of fencing per foot .

cost of fencing length 1 = $10(1)

cost of fencing breadth b = b cost of fencing per foot

cost of fencing breadth b = $12(b)

Total cost of fencing be = 10(1) + 12b

10l + 12b = 148 5l + 6b = 74 — eq 2

Step 2 :- eliminate x to find value of y

Doing 5eq1 - eq2 to eliminate x

We get 5l + 5b = (5l + 6b) = 5(14) = 74

5l + 5b - 5l - 6b = 70 - 74

-b = -4

∴ b = 4 |

∴ l = 10. |

Maths-

### A Stone of volume 36 cubic cm is placed in a rectangular tank measuring 8 cm by 6 cm by 5 cm. A student wants to fill the tank with water to a height of 4 cm. What is the volume of the water needed?

Hint:

We find the partial volume of the tank with height of the water level given. Then we subtract the volume of stone from the calculated volume of water to get the volume of water needed.

Explanations:

Step 1 of 2:

The length and width of the rectangular tank are given by, 8 cm(= ) and 6 cm(= b).

A student wants to fill the tank with water to a height of 4 cm(= h).

Then, the volume of water needed without the stone is cm

Step 2 of 2:

The volume of the stone is given by, 36 cm

Therefore, the water needed to fill up to desired height is 192 – 36 = 156 cm

Final Answer:

The water needed is 156 cm

We find the partial volume of the tank with height of the water level given. Then we subtract the volume of stone from the calculated volume of water to get the volume of water needed.

Explanations:

Step 1 of 2:

The length and width of the rectangular tank are given by, 8 cm(= ) and 6 cm(= b).

A student wants to fill the tank with water to a height of 4 cm(= h).

Then, the volume of water needed without the stone is cm

^{3}Step 2 of 2:

The volume of the stone is given by, 36 cm

^{3}Therefore, the water needed to fill up to desired height is 192 – 36 = 156 cm

^{3}Final Answer:

The water needed is 156 cm

^{3}.### A Stone of volume 36 cubic cm is placed in a rectangular tank measuring 8 cm by 6 cm by 5 cm. A student wants to fill the tank with water to a height of 4 cm. What is the volume of the water needed?

Maths-General

Hint:

We find the partial volume of the tank with height of the water level given. Then we subtract the volume of stone from the calculated volume of water to get the volume of water needed.

Explanations:

Step 1 of 2:

The length and width of the rectangular tank are given by, 8 cm(= ) and 6 cm(= b).

A student wants to fill the tank with water to a height of 4 cm(= h).

Then, the volume of water needed without the stone is cm

Step 2 of 2:

The volume of the stone is given by, 36 cm

Therefore, the water needed to fill up to desired height is 192 – 36 = 156 cm

Final Answer:

The water needed is 156 cm

We find the partial volume of the tank with height of the water level given. Then we subtract the volume of stone from the calculated volume of water to get the volume of water needed.

Explanations:

Step 1 of 2:

The length and width of the rectangular tank are given by, 8 cm(= ) and 6 cm(= b).

A student wants to fill the tank with water to a height of 4 cm(= h).

Then, the volume of water needed without the stone is cm

^{3}Step 2 of 2:

The volume of the stone is given by, 36 cm

^{3}Therefore, the water needed to fill up to desired height is 192 – 36 = 156 cm

^{3}Final Answer:

The water needed is 156 cm

^{3}.Maths-

### The area of a trapezium side field is 480sq. m , the height is 15m and one of the parallel side is 20m .Find the other side ?

Ans :- 44 m.

Explanation :-

Given, area of trapezium = 480 sq.m ,the length of parallel side is 20 m

Let the length of the other parallel side be x m .

Altitude or height = 15 m

Therefore, the length of the unknown parallel side is 44 m.

Explanation :-

Given, area of trapezium = 480 sq.m ,the length of parallel side is 20 m

Let the length of the other parallel side be x m .

Altitude or height = 15 m

Therefore, the length of the unknown parallel side is 44 m.

### The area of a trapezium side field is 480sq. m , the height is 15m and one of the parallel side is 20m .Find the other side ?

Maths-General

Ans :- 44 m.

Explanation :-

Given, area of trapezium = 480 sq.m ,the length of parallel side is 20 m

Let the length of the other parallel side be x m .

Altitude or height = 15 m

Therefore, the length of the unknown parallel side is 44 m.

Explanation :-

Given, area of trapezium = 480 sq.m ,the length of parallel side is 20 m

Let the length of the other parallel side be x m .

Altitude or height = 15 m

Therefore, the length of the unknown parallel side is 44 m.

Maths-

### The velocity v , in meters per second, of a falling object on Earth after t seconds, ignoring the effect of air resistance, is modeled by the equation v = 9.8t . There is a different linear relationship between time and velocity on Mars, as shown in the table below.

If an object dropped toward the surface of Earth has a velocity of 58.8 meters per second after t seconds, what would be the velocity of the same object dropped toward the surface of Mars after t seconds, ignoring the effect of air resistance?

First we find the equation of velocity of a falling object on Mars (ignoring the effect of air resistance).

It is given that the relationship is linear, so the ratio of velocity and time is constant.

From the table, we have

Thus, the equation of velocity with respect to time of a falling object on Mars is v = 3.7t

Given, equation of velocity of a falling object on Earth with respect to time is v = 9.8t

According to the question, an object dropped toward the surface of Earth has a velocity of 58.8 meters per second after

By the equation v = 9.8t , the time after which the velocity is 58.8 metres per second is given by

We need to find the velocity of the same object dropped towards the surface of mass after t=6 seconds

By the relation v = 3.7t , we get the velocity to be

Thus, the correct option is B)

It is given that the relationship is linear, so the ratio of velocity and time is constant.

From the table, we have

Thus, the equation of velocity with respect to time of a falling object on Mars is v = 3.7t

Given, equation of velocity of a falling object on Earth with respect to time is v = 9.8t

According to the question, an object dropped toward the surface of Earth has a velocity of 58.8 meters per second after

*t*secondsBy the equation v = 9.8t , the time after which the velocity is 58.8 metres per second is given by

We need to find the velocity of the same object dropped towards the surface of mass after t=6 seconds

By the relation v = 3.7t , we get the velocity to be

Thus, the correct option is B)

### The velocity v , in meters per second, of a falling object on Earth after t seconds, ignoring the effect of air resistance, is modeled by the equation v = 9.8t . There is a different linear relationship between time and velocity on Mars, as shown in the table below.

If an object dropped toward the surface of Earth has a velocity of 58.8 meters per second after t seconds, what would be the velocity of the same object dropped toward the surface of Mars after t seconds, ignoring the effect of air resistance?

Maths-General

First we find the equation of velocity of a falling object on Mars (ignoring the effect of air resistance).

It is given that the relationship is linear, so the ratio of velocity and time is constant.

From the table, we have

Thus, the equation of velocity with respect to time of a falling object on Mars is v = 3.7t

Given, equation of velocity of a falling object on Earth with respect to time is v = 9.8t

According to the question, an object dropped toward the surface of Earth has a velocity of 58.8 meters per second after

By the equation v = 9.8t , the time after which the velocity is 58.8 metres per second is given by

We need to find the velocity of the same object dropped towards the surface of mass after t=6 seconds

By the relation v = 3.7t , we get the velocity to be

Thus, the correct option is B)

It is given that the relationship is linear, so the ratio of velocity and time is constant.

From the table, we have

Thus, the equation of velocity with respect to time of a falling object on Mars is v = 3.7t

Given, equation of velocity of a falling object on Earth with respect to time is v = 9.8t

According to the question, an object dropped toward the surface of Earth has a velocity of 58.8 meters per second after

*t*secondsBy the equation v = 9.8t , the time after which the velocity is 58.8 metres per second is given by

We need to find the velocity of the same object dropped towards the surface of mass after t=6 seconds

By the relation v = 3.7t , we get the velocity to be

Thus, the correct option is B)

Maths-

### A pet store has 30 animals. Some are cats and the rest are dogs. The cats cost $50 each. The dogs cost $100each. If the total cost for all 30 animals is $1900, how many cats are there?

Ans :-The animal shop has 22 cats .

Explanation :-

Step 1:- construct the system of linear equations

let the no. of cats be x and let the no. of dogs be y

Given total no. of animals = no. of cats + no. of dogs = 30

So, x + y = 30 — eq1

Given total cost of all animals = cost of x cats + cost of y dogs = $1900.

cost of x cats = x cost of each cat ( The cats cost $50 each)

cost of x cats = 50x (in $ )

cost of y dogs = y cost of each dog ( The dogs cost $100each)

cost of y dogs = 100y (in $)

So total cost of all animals = 50x + 100y = 1.900 — eq2

Step 2 :- eliminate x to find value of y

Doing eq2 - 150 eq1 to eliminate x

We get 50x + 100y - 50(x + y) = 1,900 - 50(30)

Step 3:- substitute the value of y in eq1 to get the value of x

∴ The animals shop has 22 cats and 8 dogs in total of 30 animals.

Explanation :-

Step 1:- construct the system of linear equations

let the no. of cats be x and let the no. of dogs be y

Given total no. of animals = no. of cats + no. of dogs = 30

So, x + y = 30 — eq1

Given total cost of all animals = cost of x cats + cost of y dogs = $1900.

cost of x cats = x cost of each cat ( The cats cost $50 each)

cost of x cats = 50x (in $ )

cost of y dogs = y cost of each dog ( The dogs cost $100each)

cost of y dogs = 100y (in $)

So total cost of all animals = 50x + 100y = 1.900 — eq2

Step 2 :- eliminate x to find value of y

Doing eq2 - 150 eq1 to eliminate x

We get 50x + 100y - 50(x + y) = 1,900 - 50(30)

∴ y = 8 |

∴ x = 22 |

### A pet store has 30 animals. Some are cats and the rest are dogs. The cats cost $50 each. The dogs cost $100each. If the total cost for all 30 animals is $1900, how many cats are there?

Maths-General

Ans :-The animal shop has 22 cats .

Explanation :-

Step 1:- construct the system of linear equations

let the no. of cats be x and let the no. of dogs be y

Given total no. of animals = no. of cats + no. of dogs = 30

So, x + y = 30 — eq1

Given total cost of all animals = cost of x cats + cost of y dogs = $1900.

cost of x cats = x cost of each cat ( The cats cost $50 each)

cost of x cats = 50x (in $ )

cost of y dogs = y cost of each dog ( The dogs cost $100each)

cost of y dogs = 100y (in $)

So total cost of all animals = 50x + 100y = 1.900 — eq2

Step 2 :- eliminate x to find value of y

Doing eq2 - 150 eq1 to eliminate x

We get 50x + 100y - 50(x + y) = 1,900 - 50(30)

Step 3:- substitute the value of y in eq1 to get the value of x

∴ The animals shop has 22 cats and 8 dogs in total of 30 animals.

Explanation :-

Step 1:- construct the system of linear equations

let the no. of cats be x and let the no. of dogs be y

Given total no. of animals = no. of cats + no. of dogs = 30

So, x + y = 30 — eq1

Given total cost of all animals = cost of x cats + cost of y dogs = $1900.

cost of x cats = x cost of each cat ( The cats cost $50 each)

cost of x cats = 50x (in $ )

cost of y dogs = y cost of each dog ( The dogs cost $100each)

cost of y dogs = 100y (in $)

So total cost of all animals = 50x + 100y = 1.900 — eq2

Step 2 :- eliminate x to find value of y

Doing eq2 - 150 eq1 to eliminate x

We get 50x + 100y - 50(x + y) = 1,900 - 50(30)

∴ y = 8 |

∴ x = 22 |

Maths-

### The parallel sides of a trapezium are 30 cm and 10 cm and the other two sides are 13 cm and 21 cm. Find its area?

Ans :- 252 cm

Explanation :-

Step 1:- Given lengths of parallel sides is 10 and 30 cm .

Length of non parallel sides is 21 and 13 cm

We get a triangle and parallelogram ABED

we get BE = 21 cm ; DE = 10 cm (opposites sides of parallelogram )

CE = CD-DE = 30 - 10 = 20 cm .

Step 2:- Equate the areas and find value of height h

Step 3:-Find the area of trapezium

Therefore, Area of trapezium ABCD = 252 cm

^{2}.Explanation :-

Step 1:- Given lengths of parallel sides is 10 and 30 cm .

Length of non parallel sides is 21 and 13 cm

We get a triangle and parallelogram ABED

we get BE = 21 cm ; DE = 10 cm (opposites sides of parallelogram )

CE = CD-DE = 30 - 10 = 20 cm .

Step 2:- Equate the areas and find value of height h

Step 3:-Find the area of trapezium

Therefore, Area of trapezium ABCD = 252 cm

^{2}.### The parallel sides of a trapezium are 30 cm and 10 cm and the other two sides are 13 cm and 21 cm. Find its area?

Maths-General

Ans :- 252 cm

Explanation :-

Step 1:- Given lengths of parallel sides is 10 and 30 cm .

Length of non parallel sides is 21 and 13 cm

We get a triangle and parallelogram ABED

we get BE = 21 cm ; DE = 10 cm (opposites sides of parallelogram )

CE = CD-DE = 30 - 10 = 20 cm .

Step 2:- Equate the areas and find value of height h

Step 3:-Find the area of trapezium

Therefore, Area of trapezium ABCD = 252 cm

^{2}.Explanation :-

Step 1:- Given lengths of parallel sides is 10 and 30 cm .

Length of non parallel sides is 21 and 13 cm

We get a triangle and parallelogram ABED

we get BE = 21 cm ; DE = 10 cm (opposites sides of parallelogram )

CE = CD-DE = 30 - 10 = 20 cm .

Step 2:- Equate the areas and find value of height h

Step 3:-Find the area of trapezium

Therefore, Area of trapezium ABCD = 252 cm

^{2}.Maths-

### A Hollow cylindrical pipe is 21 cm long. If its outer and inner diameters are 10 cm and 6 cm respectively, then find the volume of the metal used in making the pipe?

Hint:

We find the inner and outer radius of the pipe. Then calculate the volume of the pipe using both the radius. At last, we subtract the volume with inner radius from the volume with outer radius to find the volume of metal.

Explanations:

Step 1 of 3:

The outer and inner diameters of the hollow cylindrical pipe are given by, 10 cm and 6 cm.

Then, the outer and inner radius are 5 cm(= R) and 3 cm(= r).

The pipe is 21 cm(= h) long.

Step 2 of 3:

The volume with outer radius is cm

The volume with inner radius is cm

Step 3 of 3:

Volume of metal used to make the pipe is cm

Final Answer:

The volume of metal used in making the pipe is 1056 cm

We find the inner and outer radius of the pipe. Then calculate the volume of the pipe using both the radius. At last, we subtract the volume with inner radius from the volume with outer radius to find the volume of metal.

Explanations:

Step 1 of 3:

The outer and inner diameters of the hollow cylindrical pipe are given by, 10 cm and 6 cm.

Then, the outer and inner radius are 5 cm(= R) and 3 cm(= r).

The pipe is 21 cm(= h) long.

Step 2 of 3:

The volume with outer radius is cm

^{3}The volume with inner radius is cm

^{3}Step 3 of 3:

Volume of metal used to make the pipe is cm

^{3}Final Answer:

The volume of metal used in making the pipe is 1056 cm

^{3}.### A Hollow cylindrical pipe is 21 cm long. If its outer and inner diameters are 10 cm and 6 cm respectively, then find the volume of the metal used in making the pipe?

Maths-General

Hint:

We find the inner and outer radius of the pipe. Then calculate the volume of the pipe using both the radius. At last, we subtract the volume with inner radius from the volume with outer radius to find the volume of metal.

Explanations:

Step 1 of 3:

The outer and inner diameters of the hollow cylindrical pipe are given by, 10 cm and 6 cm.

Then, the outer and inner radius are 5 cm(= R) and 3 cm(= r).

The pipe is 21 cm(= h) long.

Step 2 of 3:

The volume with outer radius is cm

The volume with inner radius is cm

Step 3 of 3:

Volume of metal used to make the pipe is cm

Final Answer:

The volume of metal used in making the pipe is 1056 cm

We find the inner and outer radius of the pipe. Then calculate the volume of the pipe using both the radius. At last, we subtract the volume with inner radius from the volume with outer radius to find the volume of metal.

Explanations:

Step 1 of 3:

The outer and inner diameters of the hollow cylindrical pipe are given by, 10 cm and 6 cm.

Then, the outer and inner radius are 5 cm(= R) and 3 cm(= r).

The pipe is 21 cm(= h) long.

Step 2 of 3:

The volume with outer radius is cm

^{3}The volume with inner radius is cm

^{3}Step 3 of 3:

Volume of metal used to make the pipe is cm

^{3}Final Answer:

The volume of metal used in making the pipe is 1056 cm

^{3}.Maths-

### There are 8 people in the lift. Few are ladies and the rest are children. Each lady weighs 150 pounds. Each child weighs 50 pounds. The total weight of the 8 people is 800 pounds. Construct the system of linear equations and use elimination method to solve.

Ans :- The no. of ladies in the lift are 4 and no. of children in the lift are 4 .

Explanation :-

Step 1:- construct the system of linear equations

let the no. of ladies in the lift be x and let the number of children be y

Given total no. of people = no. of ladies + no. of children = 8

So, x + y = 8 — eq1

Given total weight = weight of x ladies + weight of y children = 800.

weight of x ladies = x weight of each lady ( Each lady weighs 150 pounds)

weight of x ladies = 150x (pounds)

weight of y children = y weight of each child ( Each child weighs 50 pounds)

weight of y children = 50y (pounds)

So total weight = 150x + 50y = 800 — eq2

Step 2 :- eliminate x to find value of y

Doing eq2 - 150 eq1 to eliminate x

We get 150x + 50y - 150 (x + y) = 800 - 150(8)

Step 3:- substitute the value of y in eq1 to get the value of x

∴The no. of ladies in the lift are 4 and no. of children in the lift are 4 .

Explanation :-

Step 1:- construct the system of linear equations

let the no. of ladies in the lift be x and let the number of children be y

Given total no. of people = no. of ladies + no. of children = 8

So, x + y = 8 — eq1

Given total weight = weight of x ladies + weight of y children = 800.

weight of x ladies = x weight of each lady ( Each lady weighs 150 pounds)

weight of x ladies = 150x (pounds)

weight of y children = y weight of each child ( Each child weighs 50 pounds)

weight of y children = 50y (pounds)

So total weight = 150x + 50y = 800 — eq2

Step 2 :- eliminate x to find value of y

Doing eq2 - 150 eq1 to eliminate x

We get 150x + 50y - 150 (x + y) = 800 - 150(8)

∴ y = 4 |

∴ x = 4 |

### There are 8 people in the lift. Few are ladies and the rest are children. Each lady weighs 150 pounds. Each child weighs 50 pounds. The total weight of the 8 people is 800 pounds. Construct the system of linear equations and use elimination method to solve.

Maths-General

Ans :- The no. of ladies in the lift are 4 and no. of children in the lift are 4 .

Explanation :-

Step 1:- construct the system of linear equations

let the no. of ladies in the lift be x and let the number of children be y

Given total no. of people = no. of ladies + no. of children = 8

So, x + y = 8 — eq1

Given total weight = weight of x ladies + weight of y children = 800.

weight of x ladies = x weight of each lady ( Each lady weighs 150 pounds)

weight of x ladies = 150x (pounds)

weight of y children = y weight of each child ( Each child weighs 50 pounds)

weight of y children = 50y (pounds)

So total weight = 150x + 50y = 800 — eq2

Step 2 :- eliminate x to find value of y

Doing eq2 - 150 eq1 to eliminate x

We get 150x + 50y - 150 (x + y) = 800 - 150(8)

Step 3:- substitute the value of y in eq1 to get the value of x

∴The no. of ladies in the lift are 4 and no. of children in the lift are 4 .

Explanation :-

Step 1:- construct the system of linear equations

let the no. of ladies in the lift be x and let the number of children be y

Given total no. of people = no. of ladies + no. of children = 8

So, x + y = 8 — eq1

Given total weight = weight of x ladies + weight of y children = 800.

weight of x ladies = x weight of each lady ( Each lady weighs 150 pounds)

weight of x ladies = 150x (pounds)

weight of y children = y weight of each child ( Each child weighs 50 pounds)

weight of y children = 50y (pounds)

So total weight = 150x + 50y = 800 — eq2

Step 2 :- eliminate x to find value of y

Doing eq2 - 150 eq1 to eliminate x

We get 150x + 50y - 150 (x + y) = 800 - 150(8)

∴ y = 4 |

∴ x = 4 |

Maths-

### If the radius and height of a cylinder are in ratio 5 :7 and its volume is 550 cubic.cm, then find the radius?

Hint:

We use the common ratio to find the radius. Moreover, we would only use the positive of the cube root since radius is a distance and distance cannot be negative.

Explanations:

Step 1 of 3:

The radius and height of a cylinder are in ratio 5:7.

Let the common ratio be

Then, the radius and height of the cylinder is given by, 5

Step 2 of 3:

We use the common ratio to find the radius. Moreover, we would only use the positive of the cube root since radius is a distance and distance cannot be negative.

Explanations:

Step 1 of 3:

The radius and height of a cylinder are in ratio 5:7.

Let the common ratio be

*x*.Then, the radius and height of the cylinder is given by, 5

*x*(=*r*) and 7*x*(=*h*) unit, respectively.Step 2 of 3:

Step 3 of 3:

The common ratio is

Therefore, the radius of the cylinder is cm

Final Answer:

The radius is 5 cm.Step 3 of 3:

The common ratio is

*x*= 1.Therefore, the radius of the cylinder is cm

Final Answer:

The radius is 5 cm.

**If the radius and height of a cylinder are in ratio 5 :7 and its volume is 550 cubic.cm, then find the radius?**

**Maths-General**

**Hint:**

We use the common ratio to find the radius. Moreover, we would only use the positive of the cube root since radius is a distance and distance cannot be negative.

Explanations:

Step 1 of 3:

The radius and height of a cylinder are in ratio 5:7.

Let the common ratio be

Then, the radius and height of the cylinder is given by, 5

Step 2 of 3:

We use the common ratio to find the radius. Moreover, we would only use the positive of the cube root since radius is a distance and distance cannot be negative.

Explanations:

Step 1 of 3:

The radius and height of a cylinder are in ratio 5:7.

Let the common ratio be

*x*.Then, the radius and height of the cylinder is given by, 5

*x*(=*r*) and 7*x*(=*h*) unit, respectively.Step 2 of 3:

Step 3 of 3:

The common ratio is

Therefore, the radius of the cylinder is cm

Final Answer:

The radius is 5 cm.Step 3 of 3:

The common ratio is

*x*= 1.Therefore, the radius of the cylinder is cm

Final Answer:

The radius is 5 cm.

Maths-

Which of the following is true about the standard deviations of the two data sets in the table above?

The mean of data set A is given by:

The mean of data set B is given by:

We can clearly observe that the values in Data set B are dispersed from the mean, whereas the values in Data set A are not so dispersed.

Hence, the standard deviation of data set B is larger than the standard deviation of data set A.

Thus, the correct option is A

The mean of data set B is given by:

We can clearly observe that the values in Data set B are dispersed from the mean, whereas the values in Data set A are not so dispersed.

Hence, the standard deviation of data set B is larger than the standard deviation of data set A.

Thus, the correct option is A

Which of the following is true about the standard deviations of the two data sets in the table above?

Maths-General

The mean of data set A is given by:

The mean of data set B is given by:

We can clearly observe that the values in Data set B are dispersed from the mean, whereas the values in Data set A are not so dispersed.

Hence, the standard deviation of data set B is larger than the standard deviation of data set A.

Thus, the correct option is A

The mean of data set B is given by:

We can clearly observe that the values in Data set B are dispersed from the mean, whereas the values in Data set A are not so dispersed.

Hence, the standard deviation of data set B is larger than the standard deviation of data set A.

Thus, the correct option is A

Maths-

### 50 animals are there in a shed. Some of the animals have 2 legs and the rest of them have 4 legs. In total there are 172 legs. Frame the system of linear equations and use elimination method to solve .

Ans :- The no. of 2 leg animals are 26 and no. of 4 leg animals are 24.

Explanation :-

Step 1:- Frame the equations

let the no. of 2 leg animals be x and let the number of 4 leg animals be y

Given total no. of animals = no. of 2 leg animals + no. of 4 leg animals = 50

i. e x + y = 50 — eq1

Given no. of legs = 2no. of 2 leg animals + 4no. of 4 leg animals = 172.

i. e 2x+4y = 172 — eq2

So the equations are x + y = 50 and 2x + 4y = 172

Step 2 :- eliminate x to find value of y

Doing eq2 - 2eq1 to eliminate x

We get

Step 3:- substitute the value of y in eq1 to get the value of x

∴The no. of 2 leg animals are 26 and no. of 4 leg animals are 24.

Explanation :-

Step 1:- Frame the equations

let the no. of 2 leg animals be x and let the number of 4 leg animals be y

Given total no. of animals = no. of 2 leg animals + no. of 4 leg animals = 50

i. e x + y = 50 — eq1

Given no. of legs = 2no. of 2 leg animals + 4no. of 4 leg animals = 172.

i. e 2x+4y = 172 — eq2

So the equations are x + y = 50 and 2x + 4y = 172

Step 2 :- eliminate x to find value of y

Doing eq2 - 2eq1 to eliminate x

We get

∴ y = 24 |

∴ x = 26 |

### 50 animals are there in a shed. Some of the animals have 2 legs and the rest of them have 4 legs. In total there are 172 legs. Frame the system of linear equations and use elimination method to solve .

Maths-General

Ans :- The no. of 2 leg animals are 26 and no. of 4 leg animals are 24.

Explanation :-

Step 1:- Frame the equations

let the no. of 2 leg animals be x and let the number of 4 leg animals be y

Given total no. of animals = no. of 2 leg animals + no. of 4 leg animals = 50

i. e x + y = 50 — eq1

Given no. of legs = 2no. of 2 leg animals + 4no. of 4 leg animals = 172.

i. e 2x+4y = 172 — eq2

So the equations are x + y = 50 and 2x + 4y = 172

Step 2 :- eliminate x to find value of y

Doing eq2 - 2eq1 to eliminate x

We get

Step 3:- substitute the value of y in eq1 to get the value of x

∴The no. of 2 leg animals are 26 and no. of 4 leg animals are 24.

Explanation :-

Step 1:- Frame the equations

let the no. of 2 leg animals be x and let the number of 4 leg animals be y

Given total no. of animals = no. of 2 leg animals + no. of 4 leg animals = 50

i. e x + y = 50 — eq1

Given no. of legs = 2no. of 2 leg animals + 4no. of 4 leg animals = 172.

i. e 2x+4y = 172 — eq2

So the equations are x + y = 50 and 2x + 4y = 172

Step 2 :- eliminate x to find value of y

Doing eq2 - 2eq1 to eliminate x

We get

∴ y = 24 |

∴ x = 26 |

Maths-

### The area of trapezium is 384 sq.cm. If its parallel sides are in the ratio 3:5 and the perpendicular distance between them 12cm. The smaller of the parallel side is

Ans :- 24 cm

Explanation :-

Step 1:- Find the equation between x and y and find the sum of lengths of parallel sides in terms of y.

let the length of parallel sides be x and y (x<y).

Given ratio of x:y = 3:5

Sum of lengths of parallel sides =

Step 2:-substitute the values in the area formula find x

Given area of trapezium = 384 sq.cm

Altitude (or) height = 12 cm

area of trapezium = ½ × (height)(sum of lengths of parallel side)

We get x = 24 cm as x is the smaller side of parallel sides(x<y)

Therefore, the length of smaller parallel sides is 24 cm .

Explanation :-

Step 1:- Find the equation between x and y and find the sum of lengths of parallel sides in terms of y.

let the length of parallel sides be x and y (x<y).

Given ratio of x:y = 3:5

Sum of lengths of parallel sides =

Step 2:-substitute the values in the area formula find x

Given area of trapezium = 384 sq.cm

Altitude (or) height = 12 cm

area of trapezium = ½ × (height)(sum of lengths of parallel side)

We get x = 24 cm as x is the smaller side of parallel sides(x<y)

Therefore, the length of smaller parallel sides is 24 cm .

### The area of trapezium is 384 sq.cm. If its parallel sides are in the ratio 3:5 and the perpendicular distance between them 12cm. The smaller of the parallel side is

Maths-General

Ans :- 24 cm

Explanation :-

Step 1:- Find the equation between x and y and find the sum of lengths of parallel sides in terms of y.

let the length of parallel sides be x and y (x<y).

Given ratio of x:y = 3:5

Sum of lengths of parallel sides =

Step 2:-substitute the values in the area formula find x

Given area of trapezium = 384 sq.cm

Altitude (or) height = 12 cm

area of trapezium = ½ × (height)(sum of lengths of parallel side)

We get x = 24 cm as x is the smaller side of parallel sides(x<y)

Therefore, the length of smaller parallel sides is 24 cm .

Explanation :-

Step 1:- Find the equation between x and y and find the sum of lengths of parallel sides in terms of y.

let the length of parallel sides be x and y (x<y).

Given ratio of x:y = 3:5

Sum of lengths of parallel sides =

Step 2:-substitute the values in the area formula find x

Given area of trapezium = 384 sq.cm

Altitude (or) height = 12 cm

area of trapezium = ½ × (height)(sum of lengths of parallel side)

We get x = 24 cm as x is the smaller side of parallel sides(x<y)

Therefore, the length of smaller parallel sides is 24 cm .

Maths-

### A Rectangular sheet of paper 44 cm x 18 cm is rolled along its length and a cylinder is formed. What is the volume of the cylinder so formed?

Hint:

If a rectangular sheet of paper is rolled along its length to form a cylinder, then its length becomes the circumference of the cylinder base and its width becomes the height of the cylinder.

Explanations:

Step 1 of 3:

The length and width of a rectangular sheet of paper are given by, 44 cm and 18 cm, respectively.

As per the hint, if the paper is rolled along its length to form a cylinder, then the height of the cylinder would be 18 cm(=

Step 2 of 3:

Therefore, cm

The radius of the cylinder formed is 7 cm(=

Step 3 of 3:

The volume of the formed cylinder is cm

Final Answer:

The volume of the cylinder so formed is 2772 cm

If a rectangular sheet of paper is rolled along its length to form a cylinder, then its length becomes the circumference of the cylinder base and its width becomes the height of the cylinder.

Explanations:

Step 1 of 3:

The length and width of a rectangular sheet of paper are given by, 44 cm and 18 cm, respectively.

As per the hint, if the paper is rolled along its length to form a cylinder, then the height of the cylinder would be 18 cm(=

*h*), and the circumference of the cylinder base would be 44 cm.Step 2 of 3:

Therefore, cm

The radius of the cylinder formed is 7 cm(=

*r*).Step 3 of 3:

The volume of the formed cylinder is cm

^{3}Final Answer:

The volume of the cylinder so formed is 2772 cm

^{3}.### A Rectangular sheet of paper 44 cm x 18 cm is rolled along its length and a cylinder is formed. What is the volume of the cylinder so formed?

Maths-General

Hint:

If a rectangular sheet of paper is rolled along its length to form a cylinder, then its length becomes the circumference of the cylinder base and its width becomes the height of the cylinder.

Explanations:

Step 1 of 3:

The length and width of a rectangular sheet of paper are given by, 44 cm and 18 cm, respectively.

As per the hint, if the paper is rolled along its length to form a cylinder, then the height of the cylinder would be 18 cm(=

Step 2 of 3:

Therefore, cm

The radius of the cylinder formed is 7 cm(=

Step 3 of 3:

The volume of the formed cylinder is cm

Final Answer:

The volume of the cylinder so formed is 2772 cm

If a rectangular sheet of paper is rolled along its length to form a cylinder, then its length becomes the circumference of the cylinder base and its width becomes the height of the cylinder.

Explanations:

Step 1 of 3:

The length and width of a rectangular sheet of paper are given by, 44 cm and 18 cm, respectively.

As per the hint, if the paper is rolled along its length to form a cylinder, then the height of the cylinder would be 18 cm(=

*h*), and the circumference of the cylinder base would be 44 cm.Step 2 of 3:

Therefore, cm

The radius of the cylinder formed is 7 cm(=

*r*).Step 3 of 3:

The volume of the formed cylinder is cm

^{3}Final Answer:

The volume of the cylinder so formed is 2772 cm

^{3}.

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